Mathematics - Area Under Curves Question with Solution | TestHub

Intersection points of $y^2=4kx$ and $y=kx$ are obtained by substituting $y=kx$ into $y^2=4kx$.

$(kx{)}^2=4kx$

$k^2{x}^2-4kx=0$

$kx(kx-4)=0$

Since $k>0$, $x=0$ or $x=\frac{4}{k}$.

The area enclosed between the curves is given by:

$A={\int }_0^{4/k}(2\sqrt{kx}-kx)dx$

$A={\left[\frac{4}{3}\sqrt{k}{x}^{3/2}-\frac{k}{2}{x}^2\right]}_0^{4/k}$

$A=\frac{4}{3}\sqrt{k}{\left(\frac{4}{k}\right)}^{3/2}-\frac{k}{2}{\left(\frac{4}{k}\right)}^2$

$A=\frac{32}{3k}-\frac{8}{k}=\frac{8}{3k}$

Given that the area is \dfrac{2k}{3} sq. units:

$\frac{8}{3k}=\frac{2k}{3}$

$2k^2=8$

$k^2=4$

Since $k>0$, $k=2$.

Answer: $2$